Confusion with finiteness of the Legendre transform

Question

Let $f:\mathbb R \to \mathbb R$ be a convex function. In this question it is apparently proven that if $r: \mathbb R \to \mathbb R$ is such that $r(t) \stackrel{t \to \infty}{\longrightarrow} \infty$ and $f(x) \geq r(\lvert x \rvert)$ for large $x$, then the Legendre transform $$\sup_{x \in \mathbb R} \left( vx - f(x) \right)$$ is finite for all $v\in \mathbb R$. Now here is my confusion: Define the function $f(x) := \sqrt{1+x^2}$, which is (strictly) convex on $\mathbb R$. Further we obviously have $f(x) = f(\lvert x \rvert) \stackrel{x\to \infty}\longrightarrow \infty.$ But note that i.e. for $v=2$ we have $$\sup_{x\in \mathbb R} \left( 2x - \sqrt{1+x^2} \right) = \infty.$$ Where is my mistake? If I strengthen the condition on $r$ that $$\lim_{t \to \infty }\frac{r(t)}{t} = \infty$$ I think it should work.

Answer 1

The statement is clearly false, and functions which are linear at infinity are counterexamples. I seems to me that the condition is meant to be $f(x)\ge r(|x|)x$ (superlinearity, essentially) as you mentioned. If you read the proof in the post you mentioned, this is the condition that is actually used: for any $d$, $|f(x)|\ge \|d\|\|x\|$, etc.